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In the diagram shown in figure, match the following columns (take `g=10 ms^(-2)`)
image
`{:("Column I","Column II"),("(A) Normal reaction","(p) 12 SI unit"),("(B) Force of friction","(q) 20 SI unit"),("(C) Acceleration of block","(r) zero"),(,"(s) 2 SI unit"):}`

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Correct Answer - A`rarr`q,B`rarr`p,C`rarr`s
`N=mg-20sqrt(2) sin 45^(@) =20N`
`f mu_(k)N=12N`
Since, `20sqrt(2) cos 45^(@) gt muN` block will move and kinetic friction will act,
`a=(20sqrt(2) cos 45^(@)-mu_(k)N)/(m)=(20-12)/(4)=2 ms^(-2)`

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